The metal ions that have the calculated spin only magnetic moment value of 4.9 B.M.$4.9\text{ B.M.}$ are: A. Cr²⁺$\mathrm{Cr}^{2+}$ B. Fe²⁺$\mathrm{Fe}^{2+}$ C. Fe³⁺$\mathrm{Fe}^{3+}$ D. Co²⁺$\mathrm{Co}^{2+}$ E. Mn³⁺$\mathrm{Mn}^{3+}$ Choose the correct answer from the options given below:

Related Formula

Core Logic

Manganate ion (MnO₄²⁻$\mathrm{MnO}_4^{2-}$), where Mn is in +6 oxidation state, is unstable in acidic medium and undergoes disproportionation. It oxidizes to Permanganate (MnO₄^-$\mathrm{MnO}_4^-$, +7 state) and reduces to Manganese dioxide (MnO₂$\mathrm{MnO}_2$, +4 state).

Pattern Recognition

Manganate (green, +6) disproportionates in acid to Permanganate (purple, +7) and MnO₂$MnO_2$ (brown/black precipitate, +4).

Chapter Mix

Class 12 Chemistry: d and f Block Elements

Core Logic

The first reaction is the classical Chromyl Chloride Test: 4NaCl + K₂Cr₂O₇ + 6H₂SO₄ arrow 2CrO₂Cl₂ (A) + 2KHSO₄ + 4NaHSO₄ + 3H₂O$4\mathrm{NaCl} + \mathrm{K_2Cr_2O_7} + 6\mathrm{H_2SO_4} \rightarrow 2\mathrm{CrO_2Cl_2} (\text{A}) + 2\mathrm{KHSO_4} + 4\mathrm{NaHSO_4} + 3\mathrm{H_2O}$ Product A is Chromyl chloride (CrO₂Cl₂$\mathrm{CrO_2Cl_2}$), a red-orange gas.

When Chromyl chloride gas is passed into NaOH solution, it forms a yellow solution of sodium chromate (B): CrO₂Cl₂ (A) + 4NaOH arrow Na₂CrO₄ (B) + 2NaCl + 2H₂O$\mathrm{CrO_2Cl_2} (\text{A}) + 4\mathrm{NaOH} \rightarrow \mathrm{Na_2CrO_4} (\text{B}) + 2\mathrm{NaCl} + 2\mathrm{H_2O}$

Acidifying the sodium chromate solution with H₂SO₄$H_2SO_4$ and adding H₂O₂$H_2O_2$ yields a deep blue solution of Chromium(VI) peroxide, CrO₅$CrO_5$ (C): Na₂CrO₄ (B) + H₂SO₄ + 2H₂O₂ arrow CrO₅ (C) + Na₂SO₄ + 3H₂O$\mathrm{Na_2CrO_4} (\text{B}) + \mathrm{H_2SO_4} + 2\mathrm{H_2O_2} \rightarrow \mathrm{CrO_5} (\text{C}) + \mathrm{Na_2SO_4} + 3\mathrm{H_2O}$

Structure of CrO₅$CrO_5$:

• It has a butterfly structure.
• Number of peroxy units (O₂²⁻$O_2^{2-}$), X = 2$X = 2$.
• Total number of oxygen atoms, Y = 5$Y = 5$.
• Oxidation state of Cr, Z = +6$Z = +6$.

Sum: X + Y + Z = 2 + 5 + 6 = 13$X + Y + Z = 2 + 5 + 6 = 13$.

Step 1: Final Calculation

X + Y + Z = 13$X + Y + Z = 13$

Pattern Recognition

Chromyl chloride test arrow$\rightarrow$ CrO₂Cl₂$CrO_2Cl_2$ (red gas). Absorbed in NaOH arrow$\rightarrow$ Na₂CrO₄$Na_2CrO_4$ (yellow). Tested with H₂O₂/H^+$H_2O_2/H^+$ arrow$\rightarrow$ CrO₅$CrO_5$ (butterfly structure, blue, two peroxy links, Cr in +6).

Chapter Mix

Class 12 Chemistry: d and f Block Elements Class 11 Chemistry: Redox Reactions

Core Logic

• A is correct: Mn₂O₇$\text{Mn}_2\text{O}_7$ features Mn in +7 state (its highest oxidation state).
• B is correct: Oxygen stabilizes high oxidation states via multiple bonding.
• C is incorrect: Mn₂O₇$\text{Mn}_2\text{O}_7$ is a covalent green oil/oxide, not ionic.
• D is correct: Structure consists of two MnO₄$\text{MnO}_4$ tetrahedra sharing one bridging oxygen atom (O₃Mn-O-MnO₃$\text{O}_3\text{Mn}-\text{O}-\text{MnO}_3$).

Step 1: Final Conclusion

Statements A, B and D are correct, matching option (2).

Pattern Recognition

Sees: Properties and bonding of transition metal oxides like Mn₂O₇$\text{Mn}_2\text{O}_7$. Trap: Assuming high oxidation state oxides of transition metals are ionic.

Chapter Mix

Class 12 Chemistry: The d- and f-Block Elements

Related Formula

Core Logic

The first-row transition metal that does not liberate H₂$H_{2}$ gas from dilute HCl is Copper (Cu), because its standard reduction potential is positive (E°Cu²⁺/Cu = +0.34 V$E^{\circ}_{Cu^{2+}/Cu} = +0.34\text{ V}$).

When CuSO₄$CuSO_{4}$ is treated with excess KCN, it forms a very stable soluble cyano complex:

2Cu(CN)₂ arrow 2CuCN + (CN)₂$2Cu(CN)_{2} \rightarrow 2CuCN + (CN)_{2}$ CuCN + 3KCN arrow K₃[Cu(CN)₄]$CuCN + 3KCN \rightarrow K_{3}[Cu(CN)_{4}]$

The complex ion [Cu(CN)₄]³⁻$[Cu(CN)_{4}]^{3-}$ is highly stable (a perfect complex). When H₂S$H_{2}S$ is passed through this solution, it does not yield sufficient Cu⁺$Cu^{+}$ ions to exceed the solubility product (Kₛₚ$K_{sp}$) of Cu₂S$Cu_{2}S$.

Step 1: Final Conclusion

Since no copper sulphide precipitates, the amount of MS formed is 0 moles.

Pattern Recognition

Cu and Cd separation: Cu²⁺$Cu^{2+}$ forms a very stable cyanide complex that does not precipitate with H₂S$H_2S$, whereas Cd²⁺$Cd^{2+}$ forms a less stable complex that does precipitate as CdS$CdS$.

Chapter Mix

Class 12 Chemistry: d and f Block Elements Class 12 Chemistry: Coordination Compounds

Related Formula

Core Logic

Step 1: Compare IE₁$IE_1$:

Hence, IE₁(Cr) < IE₁(Mn)$IE_1(\text{Cr}) < IE_1(\text{Mn})$ (Statement-I is TRUE).

Step 2: Compare IE₂$IE_2$ and IE₃$IE_3$:

Hence, IE₃(Cr) < IE₃(Mn)$IE_3(\text{Cr}) < IE_3(\text{Mn})$. Thus Statement-II is FALSE.

Pattern Recognition

Sees: Ionization enthalpy comparison of Cr and Mn. Shortcut: Stable 3d⁵$3d^5$ configuration in Cr^+$\text{Cr}^+$ makes IE₂$IE_2$ very high, whereas 3d⁵$3d^5$ in Mn²⁺$\text{Mn}^{2+}$ makes IE₃$IE_3$ of Mn higher than Cr.

Chapter Mix

Class 12 Chemistry: d- and f-Block Elements Class 11 Chemistry: Classification of Elements and Periodicity in Properties